Datasheet

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SBOS146A − OCTOBER 1986 − REVISED AUGUST 2004
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16
DETAILED ERROR ANALYSIS
The ideal output current is:
I
O IDEAL
+ 4mA ) Ke
IN
ǒ0.016
amps
volt
)
ǒ
40
R
S
Ǔ
Ǔ
where K is the span (gain) term,
In the XTR101 there are three major components of error:
1. σ
O
= errors associated with the output stage.
2. σ
S
= errors associated with span adjustment.
3. σ
I
= errors associated with the input stage.
The transfer function including these errors is:
I
O ACTUAL
+
ǒ
4mA ) s
O
Ǔ
) K
ǒ
1 ) s
S
Ǔ
(
e
IN
) s
I
)
When this expression is expanded, second-order terms
(σ
S
, σ
I
) dropped, and terms collected, the result is:
I
O ACTUAL
+
ǒ
4mA ) s
O
Ǔ
) Ke
IN
) Ks
I
) Ks
S
e
IN
The error in the output current is I
O ACTUAL
− I
O IDEAL
and
can be found by subtracting Equation 3 from Equation 5.
I
O ERROR
+ s
O
) Ks
I
) Ks
S
e
IN
This is a general error expression. The composition of
each component of error depends on the circuitry inside
the XTR101 and the particular circuit in which it is applied.
The circuit of Figure 9 will be used to illustrate the
principles.
s
O
+ I
OS RTO
s
S
+ e
NONLINEARITY
) e
SPAN
s
I
+ V
OSI
)
ǒ
I
B1
) R
4
* I
B2
R
T
Ǔ
)
DV
CC
PSRR
)
ǒ
e
1
)e
2
Ǔ
2
* 5V
CMRR
The term in parentheses may be written in terms of offset
current and resistor mismatches as I
B1
∆R + I
OS
Ȁ R
4
.
V
OSI
(1)
= input offset voltage.
I
B1
(1)
, I
B2
(1)
= input bias current.
I
OSI
(1)
= input offset current.
I
OS RTO
(1)
= output offset current error.
∆R = R
T
− R
4
= mismatch in resistor.
∆V
CC
= change supply voltage between pins 7 and 8
away from 24V nominal.
PSRR
(1)
= power-supply rejection ratio.
CMRR
(1)
= common-mode rejection ratio.
ε
NONLIN
(1)
= span nonlinearity.
ε
SPAN
(1)
= span equation error.
Untrimmed error = 5% max. May be trimmed to zero.
(1)
These items can be found in the Electrical Characteristics.
EXAMPLE 3
See the circuit in Figure 9 with the XTR101BG
specifications and the following conditions: R
T
= 109.4Ω at
25°C, R
T
= 156.4Ω at 150°C, I
O
= 4mA at 25°C, I
O
= 20mA
at 150°C, R
S
= 123.3Ω, R
4
= 109Ω, R
L
= 250Ω,
R
LINE
= 100Ω, V
DI
= 0.6V, and V
PS
= 24V ± 0.5%.
Determine the % error at the upper and lower range
values.
A. AT THE LOWER RANGE VALUE (T = +255C)
s
O
+ I
OS RTO
+" 6mA
s
I
+ V
OSI
)
ǒ
I
BI
DR ) I
OSI
R
4
Ǔ
)
DV
CC
PSRR
)
ȧ
ȱ
Ȳ
ƪ
e
1
)e
2
ƫ
2
* 5V
CMRR
ȧ
ȳ
ȴ
DR + R
T25
o
C
* R
4
+ 109.4 * 109 [ 0
DV
CC
+
(
24 0.005
)
) 4mA
(
250W ) 100W
)
) 0.6V
+ 120mV ) 1400mV ) 600mV + 2120mV
e
1
+
ǒ
2mA 2.5kW
Ǔ
)
ǒ
1mA 109W
Ǔ
+ 5.109V
e
2
+
ǒ
2mA 2.5kW
Ǔ
)
ǒ
1mA 109.4W
Ǔ
+ 5.1094V
ǒ
e
1
) e
2
Ǔ
2
* 5V + 0.1092V
PSRR + 3.16 10
5
for 110dB
CMRR + 31.6 10
3
for 90dB
s
1
+ 30mV )
(
150nA 0 ) 20nA 109W
)
)
2120mV
3.16 10
5
)
0.1092V
3.16 10
3
+ 30mV ) 2.18mV ) 6.7mV ) 3.46mV
+ 42.34mV
s
S
+ e
NONLIN
) e
SPAN
+ 0.0001 ) 0
ǒ
assumes trim of R
S
Ǔ
I
O ERROR
+ s
O
) K s
I
) K s
S
e
IN
K + 0.016 )
40
R
S
+ 0.016 )
40
123.3W
+ 0.340
amps
volts
e
IN
+ e
2
* V
4
+ I
REF1
R
T25
o
C
* I
REF2
R
4
Since R
T 25
°
C
= R
4
:
e
IN
+
ǒ
I
REF1
* I
REF2
Ǔ
R
4
+ 0.4mA 109W
+ 43.6mV
Since the maximum mismatch of the current references is
0.04% of 1mA = 0.4µA:
I
O
error + 6mA )
ǒ
0.34AńV 42.34mV
Ǔ
)
ǒ
0.34AńV 0.0001 43.6mV
Ǔ
+ 6mA ) 14.40mA ) 0.0015mA + 20.40mA
% error +
20.40mA
16mA
100%
0.13% of span at lower range value.
(3)
(4)
(5)
(6)
(7)
(8)
(9)
(10)